A function being continuous at a domain number is the opposite of being
discontinuous.
Function values are close when the domain numbers are close.
The function \(f\) is continuous at \(a\) if
- \(a \in Dom_f\) - \(a\) is in the domain of \(f\), which means \(f(a) \in \mathbb {R}\), i.e. \(f(a)\) exists.
- For every \(\epsilon > 0\) defining an open interval around \(f(a)\), namely \(I = (f(a) - \epsilon , f(a) + \epsilon )\), there exists a corresponding \(\delta > 0\), such that the image of open interval \((a - \delta , a + \delta )\) is a subset of \((f(a)-\epsilon , f(a)+\epsilon )\).
For EVERY open interval \(I\) around \(f(a)\), there exists a corresponding open interval around \(a\), such that ALL of the function values are inside \(I\).
\(\blacktriangleright \) Discontinuity: You can find a single interval around \(f(a)\), such that EVERY interval around \(a\) contains a domain number whose function value is outside that interval.
\(\blacktriangleright \) Continuity: For any and every selected interval around \(f(a)\), you can find a corresponding interval around \(a\), such that all of their function values are inside the selected interval.
Graph of \(y=f(x)\)
![[Picture]](continuity-44d8210d76a05ee04976e7f7fdeea06e.svg)
\(f\) is continuous at \(3\).
First, \(f(3) = 2\).
Pick ANY open interval around \(2\): \((2 - \epsilon , 2 + \epsilon )\), where \(\epsilon > 0 \) is small.
Then there is a corresponding interval for the domain. Namely choose \(\delta = \frac {\epsilon }{4}\).
Let \(I = (3 - \delta , 3 + \delta ) = \left (3 - \frac {\epsilon }{4}, 3 + \frac {\epsilon }{4}\right )\)
The image of \(I\), \(f(I) = f\left (\,\left (3 - \frac {\epsilon }{4}, 3 + \frac {\epsilon }{4}\right )\,\right ) = \left (2 - \frac {\epsilon }{2}, 2 + \frac {\epsilon }{2}\right )\)
And, \(\left (2 - \frac {\epsilon }{2}, 2 + \frac {\epsilon }{2}\right ) \subset (2 - \epsilon , 2 + \epsilon ) \)
ALL of the function values land inside the original \((2 - \epsilon , 2 + \epsilon )\) interval.
This can be done FOR ANY \(\epsilon > 0\).
For ANY open interval around \(f(a)\), no matter how small, a corresponding open interval around \(a\) can be found, such that the function value from that interval around \(a\) never jump out of the interval around \(f(a)\).
Continuity: No matter what \(\epsilon \) is choosen for \(f(a)\), you can ALWAYS find a
corresponding \(\delta \) for \(a\).
Discontinuity: There is an \(\epsilon \) for \(f(a)\), such that there is no \(\delta \) for \(a\).
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more examples can be found by following this link
More Examples of Continuity